What is the general solution for -1+sin(θ)=-1 ?

The general solution for -1+sin(θ)=-1 is θ=2pin,θ=pi+2pin

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-1+sin(θ)=-1

- 1 + sin (θ) = - 1

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- 1 + sin (θ) = - 1

Add

1

to both sides

- 1 + sin (θ) + 1 = - 1 + 1

Simplify

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (x) + cos (x) = 1

2 sin^{2} (x) - sin (x) - 1 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

x, z = sin (x^{2} y)

2 tan (x) - 3 = 3 tan (x) - 4

1 + cot (θ) = csc (θ)

What is the general solution for -1+sin(θ)=-1 ?
The general solution for -1+sin(θ)=-1 is θ=2pin,θ=pi+2pin